Question

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$<b>$Show how you would connect three resistors each of resistance 6Ω, So that the combination has a resistance of (a) 9Ω (b) 4Ω ?​

Answer 1

GIVEN

Three 6Ω resistors

Total resistance is

a.9Ω

b.4Ω

TO FIND :- Circuit Arrangement

SOLUTION

a. For 9Ω resistance

First, Two 6Ω resistance should be in parallel connection and left one resistor should series to the parallel circuit.

Parallel circuit is AB and series to remaining 6Ω resistor.

Total resistance =

1/Rp = 1/6 + 1/6 = 2/6 = 1/3

Rp = 3Ω

Now it would be in series with remaining 6Ω resistor,

Rs = 3Ω + 6Ω = 9Ω (ANS)

b. For 4Ω resistance

First, Two 6Ω resistance should be in series connection and left one resistor should parallel to the series circuit.

series circuit is AB and parallel to remaining 6Ω resistor.

Rs = 6Ω + 6Ω = 12Ω

1/Rp = 1/12 + 1/6 = 3/12 = 1/4

Rp = 4Ω (ANS)

*Diagram is given for reference